Genetically engineered strains of Escherichia coli have become essential tools i

Question

Genetically engineered strains of Escherichia coli have become essential tools in the production of recombinant human peptides and proteins. One of the first substances synthesiszed using engineered E. coli was recombinant human insulin, or humulin, for the treatment of people suffering from type I diabetes mellitus. A simple reaction scheme for the production of humulin is described below. Bacteria consume glucose under aerobic conditions, and produce humulin and biomass. C_6H_12O_6 (glucose) + O_2 +NH_3 (ammonia) rightarrow C_2.3 H_2.8O_1.8 N (humulin) + CH_1.9O_0.3 N_0.3 (biomass) + CO_2 + H_2O A typical humulin production scheme consists of E. coli cultured in a large bioreactor. A continuous stream of media is fed into the reactor. A continuous stream of products and unused reactants is removed from the bioreactor for further processing, including purification of the humulin for therapeutic use. Media containing glucose and ammonia is fed into the reactor at a rate of 100 L/hr; the concentrations of glucose and ammonia in that stream are 150 mM and 50 mM, respectively. Pure oxygen gas is sparged (i.e., bubbled) into the reactor at a rate of 100 mL/min. The exit flow rate of liquid containing biomass, product, and excess reactant is 100 L/hr. Assume that there is no accumulation in the system. Assume that the reaction goes to completion. Write element balances for C, H, O, and N. Write two additional balance equations given the following information: RQ = 0.5. The ratio of humulin to biomass production is 1:5. Solve for the stoichiometric coefficients to balance the equation. Determine the inlet molar flow rates of glucose, oxygen, and ammonia in mol/hr. In this bioreactor, temperature is 310 K and pressure is 1 atm. Determine the limiting reactant, the reaction rate (R), and the fractional conversion of glucose. Calculate the molar flow rates of all constituents leaving the bioreactor. The desired output of the bioreactor is 1 kg/day of humulin. Can this output he achieved by increasing the oxygen flow rate? Justify.

Explanation / Answer

Given Data:

v0 = 100 L/hr = 1.67 L/min

Cglucose,in = 150 mM = 0.150 mol/L

Fglucose,in = 0.150 mol/L * 1.67 L/min = 0.251 mol/min

CNH3,in = 50 mM = 0.050 mol/L

FNH3,in = 0.050 mol/L * 1.67 L/min = 0.0835 mol/min

vO2= 100 mL/min = 0.10 L/min =

Mass flow rate in of O2 = 0.10 L/min * 1.429 g/L = 0.1429 g/min

FO2,in = 0.1429 g/min / 16 g/mol = 0.00893 mol/min

Now balanced equation is

C6H12O6 + 3.55 O2 + 1.3 NH3 = C2.3H2.8O1.8N + CH1.9O0.3N0.3 + 2.7 CO2 + 5.6 H2O

1 mol/min C6H12O6 = 3.55 mol/min O2 = 1.3 mol/min NH3 = 1 mol/min C2.3H2.8O1.8N = 1 mol/min CH1.9O0.3N0.3 = 2.7 mol/min CO2 = 5.6 mol/min H2O

Now we can see that oxygen is limiting reactant, hence it will consume first

to consume 0.00893 mol/min of O2:

C6H12O6 reacted = 0.00893/3.55 = 0.00252 mol/min

Fglucose,out = Fglucose,in - C6H12O6 reacted = 0.251-0.00252 =0.248 mol/min

NH3 reacted = (0.00893 * 1.3)/3.55 = 0.00327 mol/min

FNH3,out = FNH3,in - NH3 reacted = 0.0835-0.00327 =0.0802 mol/min

Humurin (C2.3H2.8O1.8N) Produced= FHumurin,out = 0.00893/3.55 = 0.00252 mol/min

Biomass (CH1.9O0.3N0.3) produced= FBiomass,out = 0.00893/3.55 = 0.00252 mol/min

CO2 produced = (0.00893*2.7)/3.55= FCO2,out = 0.0068 mol/min

H2O produced = (0.00893*5.6)/3.55= FH2O,out  = 0.0141 mol/min

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